Q 20. Profit, Loss and Discount
The increase in the price of a certain item was 25%. Then the price was decreased by 20% and then again increased by 10%. What is the resultant increase in the price?
- Let initial price = 100.
- Then is an increase in the price of 25%. So, Price after 25% increase = 125.[100+25/100]
- Then Price after 20% decrease will be 125 - (125 × 0.20) = 100.
- Again there’s a price increase. So, price after 10% increase 110.
- Thus, final value is increased by 10% with respect to the initial value (100).
Q 78. Profit, Loss and Discount
If the price of an article is decreased by 20% and then the new price is increased by 25%, then what is the net change in the price?
- Let the initial price be Rs. 100.
- New price on decreasing the original price by 20% = 100 – 20% of 100 = 100 – 20 = Rs. 80.
- Now, the final price on increasing the previous price by 25% = 80 + 25% of 80 = 80 + 20 = Rs. 100.
- So, there is no net change in price.
Q 47. Profit, Loss and Discount
A shop owner offers the following discount options on an article to a customer:
1. Successive discounts of 10% and 20%, and then pay a service tax of 10%
2. Successive discounts of 20% and 10%, and then pay a service tax of 10%
3. Pay a service tax of 10% first, then successive discounts of 20% and 10%
Which one of the following is correct?
Option D: All the options are equally good for the customer.
2. Explanation:
To determine which option is best for the customer, we need to calculate the final price the customer pays under each option. Let's assume the original price of the article is \( P \).
Option 1:
○ Apply a 10% discount:
New price = \( P \times (1 - 0.10) = 0.9P \)
○ Apply a 20% discount:
New price = \( 0.9P \times (1 - 0.20) = 0.72P \)
○ Apply a 10% service tax:
Final price = \( 0.72P \times (1 + 0.10) = 0.792P \)
Option 2:
○ Apply a 20% discount:
New price = \( P \times (1 - 0.20) = 0.8P \)
○ Apply a 10% discount:
New price = \( 0.8P \times (1 - 0.10) = 0.72P \)
○ Apply a 10% service tax:
Final price = \( 0.72P \times (1 + 0.10) = 0.792P \)
Option 3:
○ Apply a 10% service tax:
New price = \( P \times (1 + 0.10) = 1.1P \)
○ Apply a 20% discount:
New price = \( 1.1P \times (1 - 0.20) = 0.88P \)
○ Apply a 10% discount:
Final price = \( 0.88P \times (1 - 0.10) = 0.792P \)
In all three options, the final price the customer pays is \( 0.792P \). Therefore, all options are equally good for the customer.
Q 39. Profit, Loss and Discount
Rakesh had money to buy 8 mobile handsets of a specific company. But the retailer offered very good discount on that particular handset. Rakesh could buy 10 mobile handsets with the amount he had. What was the discount the retailer offered?
• Discount = Marked Price – Selling Price. Let the Marked Price of 10 articles = Rs. 100.
• So, Marked Price of 8 articles = Rs. 80. (so each was marked at Rs.10)
• So, as per the question, Rakesh purchased 10 mobile phones for Rs. 80
• Thus, he got an overall discount of Rs. 20.
• So, discount% is 20%
Q 38. Profit, Loss and Discount
A bookseller sold 'a' number of Geography textbooks at the rate of ₹x per book, 'a + 2' number of History textbooks at the rate of ₹(x - 2) per book and 'a - 2' number of Mathematics textbooks at the rate of ₹(x - 2) per book. What is his total sale in ₹?
- Sale from Geography → Rs. ax
- From History → ax + 2a + 2x + 4
- From Maths → ax – 2a – 2x + 4
- So, Total Sale is Rs. 3ax + 8.
Q 57. Profit, Loss and Discount
A shopkeeper sells an article at ₹40 and gets X% profit. However, when he sells it at ₹20, he faces same percentage of loss. What is the original cost of the article?
- Let the cost price of the article be P.
Profit % = Loss %
(40 - P)/P = (P - 20)/P
2P = 60
P = 30
- So, CP is Rs. 30.
Q 80. Profit, Loss and Discount
A person allows a 10% discount for cash payment from the marked price of a toy and still he makes a 10% gain. What is the cost price of the toy which is marked Rs. 770?
- Selling price of toy = 0.9 x marked price = 0.9 x 770
- Now SP = 1.10 CP
- CP = SP/110 = 9/11 x 770 = 630.
Q 10. Profit, Loss and Discount
What will be the expected monthly consumption (in litres) if the price goes up to ₹ 80 per litre, given the following price-consumption relationship?
Price (₹ per litre): 40, 50, 60, 75
Monthly consumption (in litres): 60, 48, 40, 32
If the price goes up to ₹ 80 per litre, his expected consumption (in litres) will be:
We know that Expenditure = Price per litre x Consumption = 40 x 60 = ₹ 2400 If the price of petrol goes up to ₹ 80 per litre. Then, consumption = 2400 / 80 = 30 L
Q 50. Profit, Loss and Discount
A person ordered 5 pairs of black socks and some pairs of brown socks. The price of a black pair was thrice that of a brown pair. While preparing the bill, the bill clerk interchanged the number of black and brown pairs by mistake which increased the bill by 100%. What was the number of pairs of brown socks in the original order?
Let the person bought x pairs of brown socks and the price of each brown pair be y
⸫ Total cost = 5 x 3y + xy
and changed cost = 5 x y + x x 3y
According to the question,
=
]
] 5y + 3xy = 15y + xy + 15y + xy
] 5y + 3xy = 30y + 2xy
] xy = 25y ] x=25 Hence, the number of pairs of brown socks is 25.
Q 77. Profit, Loss and Discount
A cow costs more than 4 goats but less than 5 goats. If a goat costs between ₹600 and ₹800, which of the following is a most valid conclusion?
Given, the cost of a goat lies between ₹ 600 to ₹ 800 and a cow costs more than 4 goats but less than 5 goats.
⸫ Minimum cost of a cow = 601 x 4 = ₹ 2404
and maximum cost of a cow = 799 x 5= ₹ 3995
So, a cow costs between ₹ 2400 and Z 4000.
Q 23. Profit, Loss and Discount
If Sohan, while selling two goats at the same price, makes a profit of 10% on one goat and suffers a loss of 10% on the other
Let the selling price of each goat be Rs. x.
CP of the goat on which there is a profit of 10%
CP of the goat on which a loss of 10% is incurred
Total cost price of the two goats = rs.
⸪ Cost price is greater than selling price
⸫ Total loss incurred =
Loss % = = 1%
Alternate Method
Loss percentage = = 1%
Hence, option (c) is correct.
Q 77. Profit, Loss and Discount
For a charity show, the total tickets sold were 420. Half of these tickets were sold at the rate of ₹5 each, one-third at the rate of ₹3 each and the rest for ₹2 each. What was the total amount received ?
Total number of tickets = 420
Tickets sold at Rs. 5 = 420/2
Tickets sold at Rs. 3 = 420/3 = 140
Remaining tickets sold at Rs. 2 = 420 - (210 + 140)
= 420 - 350= 70 tickets
Now, total amount received = (210 x 5) + (140 x 3) + (70 x 2)
=1050 + 420 + 140= 1610
Hence, option (c) is correct,
Q 62. Profit, Loss and Discount
A sum of ₹700 has to be used to give seven cash prizes to the students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, what is the least value of the prize ?
The correct answer is Option B: ₹40.
2. Explanation
To solve this problem, we need to determine the value of the smallest prize when the total sum of the prizes is ₹700, and each prize is ₹20 less than the preceding one.
Let's denote the value of the smallest prize as *x*. Therefore, the prizes can be represented as follows:
○ 1st prize: *x*
○ 2nd prize: *x + 20*
○ 3rd prize: *x + 40*
○ 4th prize: *x + 60*
○ 5th prize: *x + 80*
○ 6th prize: *x + 100*
○ 7th prize: *x + 120*
The sum of all these prizes should equal ₹700:
\[
x + (x + 20) + (x + 40) + (x + 60) + (x + 80) + (x + 100) + (x + 120) = 700
\]
Simplifying the equation:
\[
7x + (20 + 40 + 60 + 80 + 100 + 120) = 700
\]
\[
7x + 420 = 700
\]
Subtract 420 from both sides:
\[
7x = 280
\]
Divide both sides by 7:
\[
x = 40
\]
Therefore, the least value of the prize is ₹40, which corresponds to Option B.
Q 66. Profit, Loss and Discount
A person can walk a certain distance and drive back in six hours. He can also walk both ways in 10 hours. How much time will he take to drive both ways?
2. Explanation:
Let's break down the problem step by step:
○ Let \( d \) be the distance between the two points.
○ Let \( w \) be the walking speed (in distance per hour).
○ Let \( v \) be the driving speed (in distance per hour).
Given Statements:
1. A person can walk a certain distance and drive back in six hours.
2. He can also walk both ways in 10 hours.
From Statement 2:
○ Walking both ways takes 10 hours.
○ Therefore, the time to walk one way is \( \frac{10}{2} = 5 \) hours.
○ Thus, the walking speed \( w \) is \( \frac{d}{5} \).
From Statement 1:
○ Walking one way and driving back takes 6 hours.
○ Time to walk one way is 5 hours (as calculated above).
○ Therefore, the time to drive back is \( 6 - 5 = 1 \) hour.
○ Thus, the driving speed \( v \) is \( d \).
To Find:
○ The time to drive both ways.
Calculation:
○ Driving one way takes 1 hour (as calculated above).
○ Therefore, driving both ways takes \( 1 + 1 = 2 \) hours.
Hence, the time taken to drive both ways is two hours.